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Revision difference : definite integral
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The \emph{definite integral} with respect to $x$ of some function $f(x)$ over the closed interval $[a,b]$ is The \emph{definite integral} with respect to x of some function $f(x)$ over the closed interval $[a,b]$ is
defined to be the ``area under the graph of $f(x)$ with respect to $x$'' (if $f(x)$ is negative, then you have a negative area). It is written as: defined to be the ``area under the graph of $f(x)$ with respect to x'' (if f(x) is negative, then you have a negative area). It is written as:
$$ \int_a^bf(x) \ dx $$ $$ \int_a^bf(x) \ dx $$
One way to find the value of the integral is to take a limit of an approximation technique One way to find the value of the integral is to take a limit of an approximation technique
as the precision increases to infinity. as the precision increases to infinity.
For example, use a Riemann sum which approximates For example, use a Riemann Sum which approximates
the area by dividing it into $n$ intervals of equal widths, and then calculating the area the area by dividing it into $n$ intervals of equal widths, and then calculating the area
of rectangles with the width of the interval and height dependent on the function's value in the interval. of rectangles with the width of the interval and height dependent on the function's value in the interval.
Let $R_n$ be this approximation, which can be written as Let $R_n$ be this approximation, which can be written as
$$ R_n = \sum_{i=1}^{n} f(x_i^*) \Delta x $$ $$ R_n = \sum_{i=1}^{n} f(x_i^*) \Delta x $$
where $x_i^*$ is some $x$ inside the $i^{\rm th}$ interval. where $x_i^*$ is some $x$ inside the $i^{\rm th}$ interval.
Then, the integral would be Then, the integral would be
$$ \int_a^bf(x) \ dx = \lim_{n \to \infty} R_n = $$ \int_a^bf(x) \ dx = \lim_{n \to \infty} R_n =
\lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*) \Delta x $$ \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*) \Delta x $$
We can use this definition to arrive at some important properties of definite integrals We can use this definition to arrive at some important properties of definite integrals
($a$, $b$, $c$ are constant with respect to $x$): ($a$, $b$, $c$ are constant with respect to $x$):
\begin{eqnarray*} \begin{eqnarray*}
\int_a^bf(x) + g(x) \ dx & = & \int_a^bf(x)\ dx + \int_a^bg(x)\ dx \\ \int_a^bf(x) + g(x) \ dx & = & \int_a^bf(x)\ dx + \int_a^bg(x)\ dx \\
\int_a^bf(x) - g(x) \ dx & = & \int_a^bf(x)\ dx - \int_a^bg(x)\ dx \\ \int_a^bf(x) - g(x) \ dx & = & \int_a^bf(x)\ dx - \int_a^bg(x)\ dx \\
\int_a^bf(x) \ dx & = & - \int_b^af(x)\ dx \\ \int_a^bf(x) \ dx & = & - \int_b^af(x)\ dx \\
\int_a^bf(x) \ dx & = & \int_a^cf(x)\ dx + \int_c^bf(x)\ dx \\ \int_a^bf(x) \ dx & = & \int_a^cf(x)\ dx + \int_c^bf(x)\ dx \\
\int_a^bcf(x) \ dx & = & c\int_a^bf(x)\ dx \int_a^bcf(x) \ dx & = & c\int_a^bf(x)\ dx
\end{eqnarray*} \end{eqnarray*}
There are other generalisations about integrals, but many require the Fundamental Theorem of Calculus. There are other generalisations about integrals, but many require the Fundamental Theorem of Calculus.